Tag Info

Hot answers tagged

9

It is not secure, because an attacker can "mix and match" the output blocks from different authentication tags on different input messages, or repeat output blocks for repeated input blocks. For example, if the attacker knows the tag $F_k(m)$ for a one-block message $m$, then it can forge the correct tag $F_k(m) \mid F_k(m)$ for the two-block message $m ...


8

First the theoretical explanations: Integrity and authenticity are different goals to achieve, but both are achieved (for symmetric encryption) with a MAC. You should probably be using encrypt-than-MAC or an authenticated cipher unless you have very good reasons not to. No blanket statements can be made though. HMAC: HMAC is a often used construct. It ...


7

The other answer is correct in general. However, if your messages are all exactly one block long (or all one block after padding), ECB is a secure MAC. A PRP looks like a PRF up to half its bit length, i.e. up to $2^{64}$ blocks for AES. A secure PRF is a secure MAC of the same size. Thus, AES ECB used on 128-bit messages is a secure MAC as long as you use ...


6

Using a MAC on the plaintext may potentially leak information about the plaintext (MAC algorithms do not necessarily ensure confidentiality of the data they are applied to, although some MAC algorithms like HMAC seem pretty safe). If you want to avoid this (theoretical) problem, then you should encrypt the MAC on the plaintext (i.e. MAC-then-encrypt, not ...


6

Although there are already many answers here, I wanted to strongly advocate AGAINST MAC-then-encrypt. I fully agree with Thomas' first half of the answer, but completely disagree with the second half. The ciphertext is the ENTIRE ciphertext (including IV etc.), and this is what must be MACed. This is granted. However, if you MAC-then-encrypt in the ...


5

We can attack the MAC defined by: MAC(k,m)=MD5(m||k), in a chosen-messages setup, basically because MD5's collision-resistance is broken. The adversary chooses m and m' of the same length $b\ge64$ bytes, differing only in their first $\lfloor b/64\rfloor$ 64-byte blocks, such that there is a collision after hashing these blocks of m and m'. If follows that ...


5

First of all there does exist information theoretically secure message authentication codes suitable for use with a one time pad. An HMAC is not one of those information theoretically secure. As far as I recall the first article presenting such a construction is the 1981 article by Wegman and Carter: New hash functions and their use in authentication and ...


5

The pseudocode has a serious issue: changing the value of nonce2 in an otherwise valid cryptogram is not detected, and results in invalid deciphered plaintext. That would be fixed by encrypt(password, string): nonce1 := generate_random_nonce() nonce2 := generate_random_nonce() key := derive_key(nonce1, password) encrypted := nonce2 || cipher(nonce2, ...


5

NMAC is really just an "education tool" on the way to HMAC and I don't think anyone intended it to be used. The two keys are needed since the first and second hashes have different purposes. The first hash on the message is just needed to get collision resistance, whereas the second hash is supposed to provide a pseudorandom function type property. As such, ...


4

If key 2 and key 3 has a nonnegligible chance to be the same, then the attacker has a nonnegligible chance of being able to generate a valid (Message, MAC) pair. Here's how it works, if the message is not a multiple of 16, then XCBC pads the message out to the next multiple of 16; if it already is, the message remains the same. Then, XCBC logically does a ...


4

Yes, this should be secure, as it is largely compatible with KDF1 and KDF2 which basically use a 4 byte big endian encoding of the counter instead of a direct ASCII conversion to a byte. Note that this construct works fine for master keys (short length, high entropy) but may be vulnerable to length extension attacks if larger input is allowed. However, if ...


4

What you think of is called an extension attack and it turns out that this is the way to go if you would like to break the general CBC-MAC when the message length is not fixed. All that an adversary needs to do is to mount a chosen message attack. Suppose he asks for the tag on the message $m=m_1||m_2||...||m_l$. The resulting CBC MAC would be ...


4

My understanding of the term 'pepper' is that it more matches your definition 2, in that a pepper is an unknown salt, which makes it a cryptographic secret, but not a key. However, in use it is not as limited by either of your definitions: The pepper can be different (or random) for all users (like a salt). The pepper can be the same for all users (like a ...


4

In summary: Yes, HMAC is the way to go for construction of a MAC from an arbitrary concrete iterated hash. We have no constructive argument of security of the MAC constructs in the question; we even have a concrete attack when using some otherwise apparently fine hashes. I consider a hash constructed by iterating a compression function $F$ as ...


4

I'll answer in order: Output size = input size That's correct, GCM uses CTR internally. It encrypts a counter value for each block, but it only uses as many bits as required from the last block. CTR turns the block cipher into a stream cipher. IV of any size For GCM a 12 byte IV is strongly suggested as other IV lengths will require additional ...


3

It is certainly wrong to state that "MAC can only be produced with AES in CBC and CFB mode", but there seems to be a simple reason that people were inspired by these modes when thinking up possible MAC constructions: They carry along some state that incorporates information from the message while traversing the input blocks. In both modes, encrypting a block ...


3

In general (without talking about MD5): Suppose our hashfunction $H$ is a Merkle-Damgard construction using a Davies-Meyer compression function $h=(H_i,m)=E_{m_i}(H_{i-1})\oplus H_{i-1}$. Since the compression function is public, everybody is able to compute the input to the final round of the MD-Hash. In addition, if you know the input to the final round ...


3

An answer surfaced from careful reading of appropriate documentation. The MAC in the question is also defined in ANSI X9.19, and is supported by some PKCS#11 tokens as the mechanism CKM_DES3_X919_MAC_GENERAL. Other than that, this MAC can be simulated using CKM_DES_MAC_GENERAL (or CKM_DES_CBC or CKM_DES3_CBC) for all but the last block, then CKM_DES3_CBC; ...


3

Here are some advantages and disadvantages for each of the three classes of MACs, which I know about: Based on block cipher There are constructions where the security of the MAC is proven in terms of the security definition of a block cipher. This means as long as the block cipher is secure, the MAC will be secure. There are constructions where encryption ...


3

$Tag = MAC_k(\Sigma_i m_i)$. Too many attacks to enumerate. As long as the sum over the blocks remains the same, the tag remains valid. If the sum is reduced modulo $2^{\mathrm{blocksize}}$ at the end, the attacker can choose the whole message, apart from a single block used to balance the sum. $t_i = MAC_k(m_i)$ and $Tag = (t_1, ..., t_l)$. ...


3

The MAC is NOT redundant. As alluded to by PaĆ­lo Ebermann's comment, the word authentication has a different meaning in the two scenarios you mentioned. In the key exchange phase of SSH, the purpose of authentication is to ensure to both parties that they are indeed talking to the right peer (if using mutual authentication). Typically, the server ...


3

A normal security notion for MAC's is that of unforgeability. So given some set of message,tag pairs $(m_0,t_0),\ldots,(m_k,t_k)$ is should be hard to create a tag for a new message not among the $\{m_0,\ldots,m_k\}$, stated informally. In your case, you could just use $E_k(m_0)$ with the secret MAC key $k$ ($H_0$ in your notation); no need for the extra ...


3

First, terms: A MAC is a generic term for a class of cryptographic primitives. It's in the same category as "hash" or "PRNG." HMAC is a particular construction that, combined with a suitable cryptographic hash, gives a secure MAC function (it can also be used to generically refer to any HMAC algorithm, since HMAC is secure with pretty much any standard hash, ...


3

I'm going to agree with @fgrieu's marvelous post above in a back-handed way. My answer is: No, you don't have to use an HMAC. Do it anyway. As you noted, some hashes, sush as SHA-3 (especially in its Keccak form), Skein (which I was a team member on), and others will work just fine. In the case of Skein, there is a one-pass Skein-MAC that has a proof of ...


3

I would propose a rather different scheme. encrypt(password, string): nonce := generate_random_nonce() secret := pbkdf(nonce, password) mackey := kbkdf(secret, 'mackey') enckey := kbkdf(secret, 'enckey') iv := kbkdf(secret, 'iv') encrypted := cipher(iv, enckey, string) return (nonce || encrypted || mac(mackey, encrypted)) Note that I've ...


2

Actually the HMAC value is not decrypted at all. The recipient takes all the needed input and she computes the HMAC on her own side and check if the result she got it is equal to the value on the message she got. You can roughly see the HMAC algorithm as an symmetric key signature. You cannot decrypt an HMAC, you only check that the value is correct.


2

Yes, the MAC algorithm works the same way on both sides. It is a mathematical function that computes the tag from the key and the message. On the sender side, the message is processed as follows: Calculate T = MAC(key, raw_message). Send the raw_message and the tag value T. Many communication protocols append them in this order but anything will do as long ...


2

Unsurprisingly, any secure MACs are a secure choice. Assessing their relative security beyond how many bits of security they offer isn't possible in general. However, there are some differences that don't depend on the protocol: Unmodified CBC-MAC is only secure for fixed length messages, otherwise it allows some forgeries. Block cipher based MACs allow ...


2

Small addition: You do not lose integrity when using encrypt-then-MAC. Since encryption is an injection, distinct plaintexts produce distinct ciphertexts, so plaintext forgery implies ciphertext forgery, which is hard if encrypt-then-MAC is secure.


2

How about GMAC? It's a Carter-Wegman MAC that meets the requirements of being fast to compute, and parallelizable. In addition, it can be securely updated, that is, given a long message, you can compute the MAC of a modified version of that message faster than recomputing the entire MAC. For example, suppose you have a message/nonce/tag triple of $(N, M, ...



Only top voted, non community-wiki answers of a minimum length are eligible